Signals and Systems - Electrical Engineering

4.9 Response of LTI Systems to Periodic Signals 277

nExample 4.13

To illustrate the filtering of a periodic signal, consider a zero-mean pulse train

x(t)=

∑∞

k=−∞,6= 0

sin(kπ/ 2 )

kπ/ 2

ej^2 kπt

as the driving source of an RC circuit that realizes a low-pass filter (i.e., a system that tries to

keep the low-frequency harmonics and get rid of the high-frequency harmonics of the input). The

transfer function of the RC low-pass filter is

H(s)=

1

1 +s/ 100

Solution

The following script computes the frequency response of the filter at the harmonic frequencies

H(jk 0 )(see Figure 4.15).

%%%%%%%%%%%%%%%%%

% Example 4.13

%%%%%%%%%%%%%%%%

% Freq response of H(s)=1/(s/scale+1) -- low-pass filter

w0 = 2 * pi; % fundamental frequency of input

M = 20; k = 0:M - 1; w1 = k. * w0; % harmonic frequencies

H = 1./(1 + j * w1/100); Hm = abs(H); Ha = angle(H); % frequency response

subplot(211)

stem(w1, Hm, ’filled’); grid; ylabel(’—H(jω)—’)

axis([0 max(w1) 0 1.3])

subplot(212)

stem(w1, Ha, ’filled’); grid

axis([0 max(w1) -1 0])

ylabel(’¡H(jω)’); xlabel(’w (rad/sec)’)

The response due to the pulse train can be found by finding the response to each of its Fourier

series components and adding them. Approximatingx(t)usingN=20 harmonics by

xN(t)=

∑^20

k=−20,6= 0

sin(kπ/ 2 )

kπ/ 2

ej^2 kπt

Then the output voltage across the capacitor is given in the steady state,

yss(t)=

∑^20

k=−20,6= 0

H(j 2 kπ)

sin(kπ/ 2 )

kπ/ 2

ej^2 kπt